This is the first in a series of posts about the Atiyah-Singer index theorem.

Let ${V, W}$ be finite-dimensional vector spaces (over ${\mathbb{C}}$, say), and consider the space ${\hom_{\mathbb{C}}(V, W)}$ of linear maps ${T: V \rightarrow W}$. To each ${T \in \hom_{\mathbb{C}}(V, W)}$, we can assign two numbers: the dimension of the kernel ${\ker T}$ and the dimension of the cokernel ${\mathrm{coker} T}$. These are obviously nonconstant, and not even locally constant. However, the difference ${\dim \ker T - \dim \mathrm{coker} T = \dim V - \dim W}$ is constant in ${T}$.

This was a trivial observation, but it leads to something deeper. More generally, let’s consider an operator (such as, eventually, a differential operator), on an infinite-dimensional Hilbert space. Choose separable, infinite-dimensional Hilbert spaces ${V, W}$; while they are abstractly isomorphic, we don’t necessarily want to choose an isomorphism between them. Consider a bounded linear operator ${T: V \rightarrow W}$.

Definition 1 ${T}$ is Fredholm if ${T}$ is “invertible up to compact operators,” i.e. there is a bounded operator ${U: W \rightarrow U}$ such that ${TU - I}$ and ${UT - I}$ are compact.

In other words, if one forms the category of Hilbert spaces and bounded operators, and quotients by the ideal (in this category) of compact operators, then ${T}$ is invertible in the quotient category. It thus follows that adding a compact operator does not change Fredholmness: in particular, ${I + K}$ is Fredholm if ${V = W}$ and ${K: V \rightarrow V}$ is compact.

Fredholm operators are the appropriate setting for generalizing the small bit of linear algebra I mentioned earlier. In fact,

Proposition 2 A Fredholm operator ${T: V \rightarrow W}$ has a finite-dimensional kernel and cokernel.

Proof: In fact, let ${V' \subset V }$ be the kernel. Then if ${v' \in V'}$, we have $\displaystyle v' = UT v' + (I - UT) v' = (I - UT) v'$

where ${U}$ is a “pseudoinverse” to ${T}$ as above. If we let ${v'}$ range over the elements of ${v'}$ of norm one, then the right-hand-side ranges over a compact set by assumption. But a locally compact Banach space is finite-dimensional, so ${V'}$ is finite-dimensional. Taking adjoints, we can similarly see that the cokernel is finite-dimensional (because the adjoint is also Fredholm). $\Box$

The space of Fredholm operators between a pair of separable, infinite-dimensional Hilbert spaces is interesting. For instance, it has the homotopy type of ${BU \times \mathbb{Z}}$, so it is a representing space for K-theory. In particular, the space of its connected components is just ${\mathbb{Z}}$. The stratification of the space of Fredholm operators is given by the index.

Definition 3 Given a Fredholm operator ${T: V \rightarrow W}$, we define the index of ${T}$ to be ${\dim \ker T - \dim \mathrm{coker} T}$. (more…)

Back to elliptic regularity. We have a constant-coefficient partial differential operator ${P = \sum_{a: |a| \leq k} C_a D^a}$ which is elliptic, i.e. the polynomial $\displaystyle Q(\xi) = \sum_{a: |a| \leq k} C_a \xi^a$

satisfies ${|Q(\xi)| \geq \epsilon |\xi|^k}$ for ${|\xi|}$ large. We used this last property to find a near-fundamental solution to ${P}$. That is, we chose ${E}$ such that ${\hat{E} = (1-\varphi) Q^{-1}}$, where ${\varphi}$ was our arbitrary cut-off function equal to one in some neighborhood of the origin. The point of all this was that $\displaystyle P(E) = \delta - \hat{\varphi}.$

In other words, ${E}$ is near the fundamental solution. So given that ${Pf = g}$, we can use ${E}$ to “almost” obtain ${f}$ from ${g}$ by convolution ${E \ast g}$—if this were exact, we’d have the fundamental solution itself.

We now want to show that ${E}$ isn’t all that badly behaved.

The singular locus of the parametrix

We are going to show that ${\mathrm{sing} E = \{0\}}$. The basic lemma we need is the following. Fix ${m}$. Consider a smooth function ${\phi}$ such that, for each ${a}$, there is a constant ${M_a}$ with $\displaystyle |D^a \phi(x)| \leq M_a (1+|x|)^{m-|a|};$

then this is a distribution, but it is not necessarily a Schwarz function. And ${\hat{\phi}}$ cannot be expected to be one, thus. Nevertheless:

Lemma 1 ${\hat{\phi}}$ is regular outside the origin. (more…)